Acid and Base
Dissociation of diprotic acid For the solution of H 2 L H 2 L HL H a = 4.69 x 0-3 HL L - H a2 =.79 x 0-0 Assumption: since a >> a2 so the dissociation of HL is insignificant comparing with the dissociation of H 2 L, which can be treated as monoprotic acid. For the solution of L - L - H 2 O HL OH - b = w / a2 =5.48x0-5 HL H 2 O H 2 L OH - b2 = w / a =2.3x0-2
Treatment of the intermediates Systematic treatment: HL H L - HL H 2 O H 2 L OH - a2 b2 Charge balance: [H ][H 2 L ]=[L - ][OH - ] a2 = [H ][L - ]/[HL] [L - ] = [HL] a2 /[H ] b2 = [H 2 L ][OH - ]/[HL] [H 2 L ]= [HL] b2 /[OH - ] =[HL][H ] b2 / w H 2 L HL H a a b2 = w [H 2 L ]=[HL][H ]/ a
Approximations Since HL is week acid/base, the [HL] F (added) then Assume, a2 [HL]>> w, then Assume, F>> a F F H a w a a a = 2 ] [ ] [ ] [ ] [ 2 HL HL H a w a a a = F F H a a a = 2 ] [ 2 ] [ a a H = ph=/2(p a p a2 )
Polyprotic acids and bases Treatments of polyprotic acid and base are similar to that of diprotic acid and base H 3 A H 2 A - H a H 2 A - HA 2- H a2 HA 2- A 3- H a3 A 3- H 2 O HA 2- OH - b = w / a3 HA 2- H 2 O H 2 A - OH - b2 = w / a2 H 2 A - H 2 O H 3 A OH - b3 = w / a Only deal with immediate neighbors
Only deal with immediate neighbors H 3 A is treated as monoprotic acid in H 3 A solution, e.g. dissolving in water, a = a H 3 A H 2 A - H H 2 A - HA 2- H HA 2- A 3- H a a2 a3 H3A H2A - H begin F 0 0 end F-x x x a = F x 2 x a
Only deal with immediate neighbors H 2 A - and HA 2- are treated as the intermediate form for a diprotic acid. For H 2 A - For HA 2- H 3 A H 2 A - H H 2 A - HA 2- H HA 2- A 3- H H 3 A H 2 A - H H 2 A - HA 2- H [ H H 2 A - HA 2- H HA 2- A 3- H F F a2 a3 a2 w 2 [ ] for HA H ] a a2f F a2 for H A w 2 a a2 a3 a a2 a2 a3
Only deal with immediate neighbors A 3- is treated as monobasic, b = b A 3- H 2 O HA 2- OH - HA 2- H 2 O H 2 A - OH - H 2 A - H 2 O H 3 A OH - b = w / a3 b2 = w / a2 b3 = w / a A 3- H 2 O HA 2- OH - begin F 0 0 end F-x x x b b = x 2 F x
Principal species dominate type of species at certain ph At the time [H 3 A]=[H 2 A - ]= /2F H 3 A H 2 A - H a begin F 0 0 end 0.5F 0.5F x = a [ H [ H ]0.5F = 0.5F ] ph = p ph<p H 3 A is the dominate species; ph>p H 2 A - is the dominate species.
Principal species dominate type of species at certain ph
Composition of solution General form of fraction for the polyprotic acid H n A: n n n n n j A H n A H n A H H H H D D H D H D H j n n n = = = = 3 2 2 2 2 ] [ ] [ ] [ ] [ ] [ ] [ α α α
Titration
Titration Increments of reagent solution (titrant) are added to analyte until the reaction is completed Titrant (know concentration) Analyte (unknown) End point: equivalence point the quantity of added titrant is the exact amount necessary for stoichiometric reaction with the analyte Example: using known H 2 SO 4 to titrate unknown OH - H 2 SO 4 2 OH 2H 2 O 2 SO 4 at end point one mole of H 2 SO 4 reacts with 2 mole of OH.
Detection end point Detection of end point: indication chemical, electrochemical, spectroscopic means to mark change in a physical property of the solution Example: HOOC-COOH 2 MnO 4-6 H 0 CO 2 2 Mn 2 H 2 O (oxalic acid, analyte) (titrant, pink)
Terminology Titrant Analyte End point Indicator: compound with a physical property (color) that changes abruptly near the equivalence point. Titration error Blank titration Primary standard: pure and stable reagent (>99.9%) which can be accurately weighted to make titrant. Standardization: use primary standard solution to determine the non primary standard standard solution. Direct titration: titrate the analyte until end point. Back titration: add excess amount of one standard solution and use the second standard solution to titrate the excess amount of the first standard solution slow reaction with analyte or no clear indicator
Titration calculation the quantity of added titrant is the exact amount necessary for stoichiometric reaction with the analyte relate the mole of titrate to the mole of analyte Example: use 0.7344 M HCl to titrate.372 g of the mixture of Na 2 CO 3 and NaHCO 3, 29. ml of HCl was consumed. Na 2 CO 3 2HCl 2 NaCl H 2 O CO 2 NaHCO 3 HCl NaCl H 2 O CO 2 at the end point, all Na become NaCl mole Na 2 CO 3 reacts with 2 mole HCl mole NaHCO 3 reacts with mole HCl Total HCl consumed 0.7344 x 29.x0-3 =0.0238 M Assume x gram of Na 2 CO 3 in the mixture, then there is.372-x gram of NaHCO 3 Mole of Na 2 CO 3 : x/05.99(molecular weight) Mole of NaHCO 3 : (.372-x)/84.0 thus: 2 x mole of Na 2 CO 3 mole of NaHCO 3 = mole of HCl 2 x (x/05.99) (.372-x)/84.0 = 0.0238 x =0.724 g 0.724 g Na 2 CO 3 and.372-0.724 = 0.648 g NaHCO 3
Titration curve Change of the indication signal the concentration of one reactant (potential, ph, optical absorption..) with the amount of titrant Understand the chemistry during the titration The influence of the conditions to the titration e.g. ph to the sharp end point.
Types of Titration Acid-base Redox (reduction oxidation, electrochemical) Complex formation (EDTA) Precipitation
Precipitation Titration sp influence the sharpness of the end point and accuracy of precipitation titration, the larger sp the better. Ion selective electrodes are usually used to detect the change of one reactant
Example of Precipitation Titration Using AgNO 3 solution to titrate the solution containing I and Cl Titration of a mixture, less soluble precipitate (larger sp ) forms first. sp has to be sufficiently different in order to separate the two different precipitates
Example of Precipitation Titration Ag I - AgI SP (AgI)=[Ag ][I - ] [Ag ]= sp (AgI)/[I - ] [I - ] = initial concentration (unknown) [Ag] added Ag Cl - AgCl SP (AgCl)=[Ag ][Cl - ] [Ag ]= sp (AgI)/[Cl] [Cl] = initial concentration (unknown) [Ag] added-[i - ] Near the end point of I -, co precipitation may happen
Titration Curve
Buffer
Buffered Solution A buffered solution resists changes in ph when acids or bases are added or when dilution occurs. The buffer is made of the mixture of acid and its conjugate base.
What happen if a weak acid is mixed with its conjugate base? the moles of acid and conjugate base in the solution will remain close to the amounts added, because of Le Châtelier s principal. HA H A - A - H 2 O HA OH - HA dissociates very little and adding A - will make the equilibrium move to right side. A - does not react with water much and adding HA will make the equilibrium move to right side
Henderson-Hasselbalch Equation The ph of the buffer - the central equation to treat buffer HA H A - a a = [H ][A - ]/[HA] log a =log[h ]log{[a - ]/[HA]} ph = p a log{[a - ]/[HA]}
The same relation, different expression A - H 2 O HA OH - b b = [HA][OH - ]/[A - ] log b = log [OH - ] log [HA]/[A - ] [OH - ]= w /[H ] log b = log w log[h ] log [HA]/[A - ] -log[h ]=log b log w - log [HA]/[A - ] ph = -log w / b log [A - ]/[HA] ph = p a log [A - ]/[HA]
Diprotic Buffers A buffer made from a diprotic (H 2 L) or polyprotic (H n L) acid a H 2 L HL - H HL - L 2- H a2 Two Henderson-Hasselbalch equations, they are both true in the same solution ph=p a log([hl - ]/[H 2 L]) ph=p a2 log([l 2- ]/[HL - ])
The ph change in buffer ph = p a log{[a - ]/[HA]}